How do you factor #36x ^ { 2} + 113x + 72#?

1 Answer
George C.
Apr 6, 2017

#36x^2+113x+72 = (4x+9)(9x+8)#

Explanation:

Given:

#36x^2+113x+72#

Use an AC method:

Look for a pair of factors of #AC=36*72=2^5*3^4# with sum #B=113#

Note that #113# is not divisible by #2# or #3#. Hence all of the prime factors #2# must be in one of the pair and all of the factors #3# must be in one of the pair.

Try:

#2^5+3^4=32+81 = 113" "# ... Yes!

So use the pair #2^5=32# and #3^4=81# to split the middle term, and factor by grouping:

#36x^2+113x+72 = 36x^2+32x+81x+72#

#color(white)(36x^2+113x+72) = (36x^2+32x)+(81x+72)#

#color(white)(36x^2+113x+72) = 4x(9x+8)+9(9x+8)#

#color(white)(36x^2+113x+72) = (4x+9)(9x+8)#

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